Difference between revisions of "2010 AIME II Problems/Problem 6"
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==Solution== | ==Solution== | ||
− | + | There are 2 ways for a monic fourth degree polynomial to be factored, into a cubic and a linear equation, or 2 quadratics. | |
+ | |||
+ | <br/> | ||
+ | <b>Case 1)</b> cubic and linear | ||
+ | |||
+ | Let <math>(x-r_1)</math> be the linear equation (it must contain one root of the quatic) | ||
+ | |||
+ | and <math>x^3+ax^2+bx+c</math> be the cubic. | ||
+ | |||
+ | By rational roots theorem, <math>r_1=1,3,7, 9</math>, or <math>63</math> | ||
+ | |||
+ | <math>(x^3+ax^2+bx+c)(x-r_1)=x^4+(a-r_1)x^3+(b-ar_1)x^2+(c-br_1)x-cr_1</math> | ||
+ | |||
+ | So <math>a-r_1=0 \rightarrow a=r_1</math>, <math>b-ar_1=0\rightarrow b=a^2</math>, | ||
+ | |||
+ | <math>c-br_1=-n\rightarrow n=a^3-c</math>, and <math>-cr_1=63 \rightarrow c=\frac{-63}{a}</math>. | ||
+ | |||
+ | <br/> | ||
+ | |||
+ | So <math>n=a^3+\frac{63}{a}</math>, <math>r_1=1,3,7, 9</math>, or <math>63</math>, which reach minimum when <math>r_1=3</math>, where <math>n=48</math> | ||
+ | |||
+ | <br/> | ||
+ | |||
+ | <br/> | ||
+ | <b>Case 2)</b> 2 quadratic | ||
+ | |||
+ | Let <math>x^2+ax+b</math> and <math>x^2+cx+d</math> be the two quadratics, | ||
+ | |||
+ | <math>(x^2 + ax + b )(x^2 + cx + d) = x^4 + (a + c)x^3 + (b + d + ac)x^2 + (ad + bc)x + bd</math> | ||
+ | |||
+ | Therefore, we have | ||
+ | |||
+ | <math>a + c = 0\rightarrow a=-c</math>, <math>b + d + ac = 0\rightarrow b+d=a^2</math> , <math>ad + bc = - n</math> | ||
+ | |||
+ | and <math>bd = 63</math>. | ||
+ | |||
+ | <math>b+d=a^2</math> ,hence the only possible values for (b,d) are (1,63) and (7,9). From this we find that the possible values for n are <math>\pm 8 * 62</math> and <math>\pm 4 * 2</math>. Therefore, the answer is <math>\boxed{8}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2010|num-b=5|num-a=7|n=II}} | {{AIME box|year=2010|num-b=5|num-a=7|n=II}} |
Revision as of 18:09, 3 April 2010
Problem
Find the smallest positive integer n with the property that the polynomial can be written as a product of two nonconstant polynomials with integer coefficients.
Solution
There are 2 ways for a monic fourth degree polynomial to be factored, into a cubic and a linear equation, or 2 quadratics.
Case 1) cubic and linear
Let be the linear equation (it must contain one root of the quatic)
and be the cubic.
By rational roots theorem, , or
So ,
,
, and
.
So ,
, or
, which reach minimum when
, where
Case 2) 2 quadratic
Let and
be the two quadratics,
Therefore, we have
,
,
and .
,hence the only possible values for (b,d) are (1,63) and (7,9). From this we find that the possible values for n are
and
. Therefore, the answer is
.
See also
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |